A) Farsightedness, 40 cm
B) Nearsightedness, 40 cm
C) Astigmatism, 40 cm
D) Nearsightedness, 250 cm
Correct Answer: B
Solution :
Negative power is given, so defect of eye is nearsigntedness Also defected far point \[=-f=-\frac{1}{p}=-\frac{100}{(-2.5)}=40\,cm\]You need to login to perform this action.
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